151 Shortcut Maths Tricks for Competitive Exams

0.5555 + 0.555 + 0.55 + 0.5 = ?
To find the sum of those number in which one number is repeated after decimal, then first write the number in either increasing
or decreasing order. Then -find the sum by using the below method.
0.5555 + 0.555 + 0.55 + 0.5
= 5 × 4 / 5 × 3 / 5 × 2 / 5 × 1
= 20 / 15 / 10 / 5 = 2.1605
5 Those numbers whose all digits are 3.
(33)2 = 1089 Those number. in which all digits are number is 3 two or more than 2 times repeated, to find the square of
these number, we repeat 1 and 8 by (n – 1) time. Where n = Number of times 3 repeated.
(333)2 = 110889
(3333)2 = 11108889

Those number whose all digits are 9.
(99)2 = 9801
(999)2 = 998001
(9999)2 = 99980001
(99999)2 = 9999800001

Those number whose all digits are 1.
A number whose one’s, ten’s, hundred’s digit is 1 i.e., 11, 111, 1111, ....
In this we count number of digits. We write 1, 2, 3, ..... in their square the digit in the number, then write in decreasing order up to1.
112 = 121
1112 = 12321
11112 = 1234321

(i) Complete square of a no. is possible if its last digit is 0, 1, 4, 5, 6 & 9. If last digit of a no. is 2, 3, 7, 8 then complete square root
of this no. is not possible.

(ii) If last digit of a no. is 1, then last digit of its complete square root is either 1 or 9.

(iii) If last digit of a no. is 4, then last digit of its complete square root is either 2 or 8.

(iv) If last digit of a no. is 5 or 0, then last digit of its complete square root is either 5 or 0.

(v) If last digit of a no. is 6, then last digit of its complete square root is either 4 or 6.

(vi) If last digit of a no. is 9, then last digit of its complete square root is either 3 or 7.

(i) Find the approx square root of given no. Divide the given no. by the prime no. less than approx square root of no. If given
no. is not divisible by any of these prime no. then the no. is prime otherwise not.
For example : To check 359 is a prime number or not.
Sol. Approx sq. root = 19
Prime no.

(ii) There are 15 prime no. from 1 to 50.

(iii) There are 25 prime no. from 1 to 100.

(iv) There are 168 prime no. from 1 to 1000.

If a no. is in the form of xn + an, then it is divisible by (x + a); if n is odd.

If xn ÷ (x – 1), then remainder is always 1.

(i) If n is even, then remainder is 1.

(ii) If n is odd, then remainder is x.
4P+1+1

(i) Value of P+ P+ P+..........¥ =
2
4P+1-1

(ii) Value of P- P- P-..........¥ =
2

(iii) Value of P. P. P...........¥ =P
( 2n-1 ) ÷2n

(iv) Value of P P P P P =P
[Where n = no. of times P repeated].
Note: If factors of P are n & (n + 1) type then value of P+ P+ P+....¥ =(n+1) and P- P- P-....¥ =n.

(i) If N is any no. and N = an × bm × cp × .... where a, b, c are prime no.
No. of divisors of N = (n + 1) (m + 1) (p + 1) ....
e.g. Find the no. of divisors of 90000.
N = 90000 = 22 × 32 × 52 × 102 = 22 × 32 × 52 × (2 × 5)2 = 24 × 32 × 54
So, the no. of divisors = (4 + 1) (2 + 1) (4 + 1) = 75

(ii) N = an × bm × cp, where a, b, c are prime
Then set of co-prime factors of N = [(n + 1) (m + 1) (p + 1) – 1 + nm + mp + pn + 3mnp]
( an+1 -1 )( bm+1 -1 )( cp+1 -1 )

(iii) If N = an × bm × cp..., where a, b & c are prime no. Then sum of the divisors =
(a-1)(b-1)(c-1)

(i) If the last digit or digit at the unit’s place of a is 1, 5 or 6, whatever be the value of n, it will have the same digit at unit’s place,
i.e.,
(.....1)n =(........1)
(.....5)n =(........5)
(.....6)n =(........6)

(ii) If the last digit or digit at the units place of a is 2, 3, 5, 7 or 8, then the last digit of an depends upon the value of n and follows
a repeating pattern in terms of 4 as given below :
n last digit of (....2)n last digit of (....3)n last digit of (....7)n last digit of (....8)n
4x+1 2 3 7 8
4x+2 4 9 9 4
4x+3 8 7 3 2
4x 6 1 1 6

(iii) If the last digit or digit at the unit’s place of a is either 4 or 9, then the last digit of an depends upon the value of n and follows
repeating pattern in terms of 2 as given below.
n last digit of (....4)n last digit of (....9)n
2x 6 1
2x + 1 4 9
(n)(n+1)

(i) Sum of n natural number =
2

(ii) Sum of n even number = (n) (n + 1)

(iii) Sum of n odd number = n2
n(n+1)(2n+1)

(i) Sum of sq. of first n natural no. =
6
n ( 4n2 -1 )

(ii) Sum of sq. of first n odd natural no. =
3
2n(n+1)(2n+1)

(iii) Sum of sq. of first n even natural no. =
3
n2(n+1)2 én(n+1)ù 2

(i) Sum of cube of first n natural no. = =ê ú
4 ë 2 û

(ii) Sum of cube of first n even natural no. = 2n2 (n + 1)2

(iii) Sum of cube of first n odd natural no. = n2 (2n2 – 1)

(i) xn – yn is divisible by (x + y)
When n is even

(ii) xn – yn is divisible by (x – y)
When n is either odd or even.

For any integer n, n3 – n is divisible by 3, n5 – n is divisible by 5, n11 – n is divisible by 11, n13 – n is divisible by 13.

(i) A no. of 3-digits which is formed by repeating a digit 3-times, then this no. is divisible by 3 and 37.
e.g., 111, 222, 333, .......

(ii) A no. of 6-digit which is formed by repeating a digit 6-times then this no. is divisible by 3, 7, 11, 13 and 37.
e.g., 111111, 222222, 333333, 444444, .............

Divisible by 7 : We use osculator (– 2) for divisibility test.
99995 : 9999 – 2 × 5 = 9989
9989 : 998 – 2 × 9 = 980
980 : 98 – 2 × 0 = 98
Now 98 is divisible by 7, so 99995 is also divisible by 7.

Divisible by 11 : In a number, if difference of sum of digit at even places and sum of digit at odd places is either 0 or multiple of
11, then no. is divisible by 11.
For example, 12342 ÷ 11
Sum of even place digit = 2 + 4 = 6
Sum of odd place digit = 1 + 3 + 2 = 6
Difference = 6 – 6 = 0
\ 12342 is divisible by 11.

Divisible by 13 : We use (+ 4) as osculator.
e.g., 876538 ÷ 13
876538: 8 × 4 + 3 = 35
5 × 4 + 3 + 5 = 28
8 × 4 + 2 + 6 = 40
0 × 4 + 4 + 7 = 11
1 × 4 + 1 + 8 = 13
13 is divisible by 13.
\ 876538 is also divisible by 13.

Divisible by 17 : We use (– 5) as osculator.
e.g., 294678: 29467 – 5 × 8 = 29427
27427: 2942 – 5 × 7 = 2907
2907: 290 – 5 × 7 = 255
255: 25 – 5 × 5 = 0
\ 294678 is completely divisible by 17.

Divisible by 19 : We use (+ 2) as osculator.
e.g: 149264: 4 × 2 + 6 = 14
4 × 2 + 1 + 2 = 11
1 × 2 + 1 + 9 = 12
2 × 2 + 1 + 4 = 9
9 × 2 + 1 = 19
19 is divisible by 19
\ 149264 is divisible by 19.

(a) Factor method (b) Division method

(i) For two no. a and b if a

(ii) The greatest number by which x, y and z completely divisible is the HCF of x, y and z.

(iii) The greatest number by which x, y, z divisible and gives the remainder a, b and c is the HCF of (x –a), (y–b) and (z–c).

(iv) The greatest number by which x, y and z divisible and gives same remainder in each case, that number is HCF of (x–y),
(y–z) and (z–x).
a c e H.C.M. of (a, c, e)

(v) H.C.F. of , and =
b d f L.C.M. of (b, d, f)

(a) Factor method (b) Division method

(i) For two numbers a and b if a

(ii) If ratio between two numbers is a : b and their H.C.F. is x, then their L.C.M. = abx.

x

(iii) If ratio between two numbers is a : b and their L.C.M. is x, then their H.C.F. =
ab

(iv) The smallest number which is divisible by x, y and z is L.C.M. of x, y and z.

(v) The smallest number which is divided by x, y and z give remainder a, b and c, but (x – a) = (y – b) = (z – c) = k, then number
is (L.C.M. of (x, y and z) – k).

(vi) The smallest number which is divided by x, y and z give remainder k in each case, then number is (L.C.M. of x, y and z) +k.
a c e L.C.M. of (a, c, e)

(vii) L.C.M. of , and =
b d f H.C.F. of (b, d, f)

(viii) For two numbers a and b –
LCM × HCF = a × b

(ix) If a is the H.C.F. of each pair from n numbers and L is L.C.M., then product of n numbers = an–1.L
ALGEBRA

(i) (a + b)2 + (a – b)2 = 2 (a2 + b2) (ii) (a + b)2 – (a – b)2 = 4ab

(iii) a3 + b3 = (a + b) (a2 – ab + b2) (iv) a3 – b3 = (a – b) (a2 + ab + b2)

(v) a4 + a2 + 1 = (a2 + a + 1) (a2 – a + 1) (vi) If a + b + c = 0, then a3 + b3 + c3=3abc
(a+b)2 -(a-b)2 (a+b)2 +(a-b)2

(vii) =4 (viii) =2
ab a2 +b2
b e h k æb e h kö

(ix) a +d +g -j =(a+d+g-j)+ç + + - ÷
c f i l èc f i l ø

(x) If a + b + c = abc, then
æ 2a ö æ 2b ö æ 2c ö æ 2a ö æ 2b ö æ 2c ö
ç ÷+ç ÷+ç ÷ = ç ÷ . ç ÷ . ç ÷ and
è1-a2
ø
è1-b2
ø
è1-c2
ø
è1-a2ø è1-b2ø è1-c2ø
æ3a-a3ö æ3b-b3ö æ3c-c3ö æ3a-a3ö æ 3b-b3ö æ3c-c3ö
ç ç è1-3a2 ÷ ÷
ø
+ç ç è1-3b2 ÷ ÷
ø
+ç ç è1-3c2 ÷ ÷
ø
= ç è1-3a2 ÷
ø
.ç
è
- 1-3b2 ÷
ø
.ç è1-3c2 ÷
ø

(i) If 1 ¹ 1 , one solution. (ii) If 1 = 1 = 1 , Infinite many solutions.
a b a b c
2 2 2 2 2
a b c

(iii) If 1 = 1 ¹ 1 , No solution
a b c
2 2 2
1 1

If a and b are roots of ax2 + bx + c = 0, then and are roots of cx2 + bx + a = 0
a b

(i) One root is zero if c = 0.

(ii) Both roots zero if b = 0 and c = 0.

(iii) Roots are reciprocal to each other, if c = a.

(iv) If both roots a and b are positive, then sign of a and b are opposite and sign of c and a are same.

(v) If both roots a and b are negative, then sign of a, b and c are same.
b c
(a+b)=- ,ab= , then
a a
a-b=
(a+b)2
-4ab
a 4 +b 4 = ( a 2 +b 2 )2 -2a 2 b 2 = é(a+b)2 -2abù 2 -2(ab)2
ë û

(i) If a, a + d, a + 2d, ..... are in A.P., then, nth term of A.P. a = a + (n – 1)d
n
n n
Sum of n terms of this A.P. = S n = é ë 2a+(n-1)dù û = [a+l] where l = last term
2 2
a = first term
d = common difference
a+b

(ii) A.M. = [ A.M. = Arithmetic mean]
Q
2

(i) G.P. = a, ar, ar2,.........
Then, nth term of G.P. a = arn–1
n
a ( rn -1 )
S = ,r>1
n (r-1)
a(1-rn)
= ,r

(ii) G.M.= ab
1 1 1

(i) A.M. × H.M. = G.M.2

(ii) A.M. > G.M. > H.M.
A.M. = Arithmetic Mean
G.M. = Geometric Mean
H.M. = Harmonic Mean
AVERAGE
n+1

(i) Average of first n natural no. =
2

(ii) Average of first n even no. = (n + 1)

(iii) Average of first n odd no. = n
(n+1)(2n+1)

(i) Average of sum of square of first n natural no. =
6
2(n+1)(2n+1)

(ii) Average of sum of square of first n even no. =
3
æ4n2 -1ö

(iii) Average of sum of square of first odd no. = ç ç 3 ÷ ÷
è ø

n(n+1)2

(i) Average of cube of first n natural no. =
4

(ii) Average of cube of first n even natural no. = 2n(n + 1)2

(iii) Average of cube of first n odd natural no. = n(2n2 – 1)
m(n+1)

Average of first n multiple of m =
2

(i) If average of some observations is x and a is added in each observations, then new average is (x + a).

(ii) If average of some observations is x and a is subtracted in each observations, then new average is (x – a).

(iii) If average of some observations is x and each observations multiply by a, then new average is ax.
x

(iv) If average of some observations is x and each observations is divided by a, then new average is .
a
n A +n A

(v) If average of n is A , & average of n is A , then Average of (n + n ) is 1 1 2 2 and
1 1 2 2 1 2 n +n
1 2
n A -n A
Average of (n – n ) is 1 1 2 2
1 2 n -n
1 2

When a person is included or excluded the group, then age/weight of that person = No. of persons in group × (Increase /
Decrease) in average ±New average.
For example : In a class average age of 15 students is 18 yrs. When the age of teacher is included their average increased by 2
yrs, then find the age of teacher.
Sol. Age of teacher = 15 × 2 + (18 + 2) = 30 + 20 = 50 yrs.

When two or more than two persons included or excluded the group, then average age of included or excluded person is
No.of person´(Increase/Decrease)inaverage±New average´(No.of personincludedorexcluded)
=
No.of includedorperson
For example : Average weight of 13 students is 44 kg. After including two new students their average weight becomes 48 kg, then
find the average weight of two new students.
Sol. Average weight of two new students
13´(48-44)+48´2 13´4+48´2 52+96
= = = = 74 kg
2 2 2
2xy

If a person travels two equal distances at a speed of x km/h and y km/h, then average speed = km/h
x+y
3xyz

If a person travels three equal distances at a speed of x km/h, y km/h and z km/h, then average speed = km/h.
xy+yz+zx
RATIO & PROPORTION
a b c a+b+c+.... K +K +K +.....

(i) If = = =.... , then = 1 2 3
K K K c K
1 2 3 3
P Q R P+Q+R
For example: If = = , then find
3 4 7 R
Sol. P = 3, Q = 4, R = 7
P+Q+R 3+4+7
Then = =2
R 7
a a a a a

(ii) If 1 = 2 = 3 = 4 =.... n =K, then a : a = (K)n
a a a a a 1 n + 1
2 3 4 5 n+1

ad-bc

A number added or subtracted from a, b, c & d, so that they are in proportion = (a+d)-(b+c)
For example : When a number should be subtracted from 2, 3, 1 & 5 so that they are in proportion. Find that number.
2´5-3´1 10-3 7
= =
Sol. Req. No. =(2+5)-(3+1)
7-4 3

If X part of A is equal to Y part of B, then A : B = Y : X.
For example: If 20% of A = 30% of B, then find A : B.
30% 3
Sol. A : B = = = 3 : 2
20% 2

When Xth part of P, Yth part of Q and Zth part of R are equal, then find A : B : C.
Then, A : B : C = yz : zx : xy
TIME, DISTANCE AND WORK
t t

A can do a/b part of work in t days and c/d part of work in t days, then 1 = 2
1 2 a/b c/d

(i) If A is K times efficient than B, Then T(K + 1) = Kt
B

(ii) If A is K times efficient than B and takes t days less than B
Kt t t
Then T = or , t = = kt
K2 -1 K-1 B K-1 A

(i) If a cistern takes X min to be filled by a pipe but due to a leak, it takes Y extra minutes to be filled, then the time taken by leak
æX2 +XYö
to empty the cistern =ç ÷min
ç Y ÷
è ø

(ii) If a leak empty a cistern in X hours. A pipe which admits Y litres per hour water into the cistern and now cistern is emptied
æX+Y+Zö
in Z hours, then capacity of cistern is = ç ÷litres.
è Z-X ø

(iii) If two pipes A and B fill a cistern in x hours and y hours. A pipe is also an outlet C. If all the three pipes are opened together,
é xyT ù
the tank full in T hours. Then the time taken by C to empty the full tank is = ê ú
ëyT+xT-xyû

(i) If t and t time taken to travel from A to B and B to A, with speed a km/h and b km/h, then distance from A to B is
1 2
æ ab ö æ ab ö
d=(t 1 +t 2 ) ç ÷ d=(t 1 -t 2 ) ç ÷
èa+bø èa-bø
æ t t ö
d=(a-b)
ç
1 2
÷
èt 1 -t 2 ø

(ii) If Ist part of distance is covered at the speed of a in t time and the second part is covered at the speed of b in t time, then
1 2
æat +bt ö
2 1
the average speed =
ç ÷
è t +t ø
1 2

PERCENTAGE

Simple Fraction Their Percentage Simple Fraction Their Percentage
1 100% 1
12.5%
8
1
50%
2
1
11.11%
1 9
33.3%
3
1
10%
1 10
25%
4
1
1 9.09%
20% 11
5
1
1 8.33%
16.67% 12
6
1
14.28%
7
æ aö æ a ö

(i) If A is çx%= ÷ more than B, then B is ç %÷ less than A.
è bø èa+b ø
æ aö æ a ö

(ii) If A is çx%= ÷less than B, then B is ç %÷more than A
è bø èa-b ø
if a > b, we take a – b
if b > a, we take b – a.
æb-a ö

If price of a article increase from ` a to ` b, then its expenses decrease by ç ´100÷%so that expenditure will be same.
è b ø

Due to increase/decrease the price x%, A man purchase a kg more in ` y, then
æ xy ö
Per kg increase or decrease = ç ÷
è100´aø
xy
Per kg starting price = `(100±x)a

For two articles, if price:
Ist IInd Overall
æ xy ö
Increase (x%) Increase (y%) Increase çx+y+ ÷%
è 100ø
æ xy ö
Increase (x%) Decrease (y%) çx-y- ÷%
è 100ø
If +ve (Increase)
If –ve (Decrease)
æ xy ö
Decrease (x%) Decrease (y%) Decreaseçx+y- ÷%
è 100ø
æ x2 ö
Increase (x%) Decrease (x%) Decrease ç ç100 ÷ ÷ %
è ø
æ x2 ö

If the side of a square or radius of a circle is x% increase/decrease, then its area increase/decrease = ç ç 2x± 100 ÷ ÷ %
è ø

If the side of a square, x% increase/decrease then x% its perimeter and diagonal increase/decrease.
t
æ100±Rö

(i) If population P increase/decrease at r% rate, then after t years population = Pç ÷
è 100 ø

(ii) If population P increase/decrease r % first year, r % increase/decrease second year and r % increase/decrease third year,
1 2 3
æ r öæ r öæ r ö
then after 3 years population = P 1± 1 1± 2 1± 3
ç ÷ç ÷ç ÷
è 100øè 100øè 100ø
If increase we use (+), if decrease we use (–)

If a man spend x% of this income on food, y% of remaining on rent and z% of remaining on cloths. If he has ` P remaining, then
P´100´100´100
total income of man is =
(100-x)(100-y)(100-z)
[Note: We can use this table for area increase/decrease in mensuration for rectangle, triangle and parallelogram].
PROFIT AND LOSS

If CP of x things = SP of y things, then
éx-y ù
Profit/Loss = ê ´100ú%
ë y û
If +ve, Profit;
If –ve, Loss

If after selling x things P/L is equal to SP of y things,
y
then P/L = ´100
(x±y)
éProfit=-ù
ê ú
ëLoss=+ û

If CP of two articles are same, and they sold at
Ist IInd Overall
æx+yö
(x%) Profit (y%) Profit ç ÷% Profit
è 2 ø
æx-yö ìProfit,if x>y
(x%) Profit (y%) Loss ç ÷%í
è 2 ø îLoss,if x

If SP of two articles are same and they sold at
Ist IInd Overall
æ x2 ö
Profit (x%) Loss(x%) Lossç ç100 ÷ ÷ %
è ø
æ100(x-y)-2xyö é2(100+x)(100-y) ù ìIf +ve,thenProfit%
Profit (x%) Loss (y%) ç è 200+x-y ÷ ø %or ê ë 200+x-y -100ú û %í îIf -ve,thenLoss%
é P+D ù

After D% discount, requires P% profit, then total increase in C.P.= ê ´100 ú %
ë100-D û
(100+P)

M.P. = C.P ×
(100-D)
(M.P.-C.P.)´100

Profit % =
C.P.

(i) For discount r 1 % and r 2 %, successive discount =ê ë
éæ
ç è
10
1
0
0
+
0
r 1ö
÷ ø
æ
ç è
10
1
0
0
+
0
r 2ö
÷ ø
æ
ç è
10
1
0
0
+
0
r 3ö
÷ ø -1
ù
ú û ´100

(ii) For discount r 1 %, r 2 % and r 3 %, successive discount =ê ë
éæ
ç è
10
1
0
0
+
0
r 1ö
÷ ø
æ
ç è
10
1
0
0
+
0
r 2ö
÷ ø
æ
ç è
10
1
0
0
+
0
r 3ö
÷ ø -1
ù
ú û ´100
SIMPLE AND COMPOUND INTEREST
If P = Principal, R = Rate per annum,
T = Time in years, SI = Simple interest,
A = Amount
PRT

(i) SI=
100
é RTù

(ii) A=P+SI=P 1+
ê ú
ë 100û

If P = Principal, A = Amount
in n years, R = rate of interest per annum.
n
é R ù
A=P 1+ , interest payable annually
ê ú
ë 100û
é R¢ ù
n¢

(i) A=P 1+ , interest payable half-yearly
ê ú
ë 100û
R¢ = R/2, n¢ = 2n
4n
é R ù

(ii) A=P 1+ , interest payable quarterly;
ê ú
ë 400û
é R ù

(i) ê 1+ úis the yearly growth factor;
ë 400û
é R ù

(ii) ê 1– úis the yearly decay factor or depreciation factor..
ë 400û
3

When time is fraction of a year, say 4 , years, then,
4
é 3 ù
é R ù 4 ê 4 R ú
Amount =P ê 1+ ú ´ê1+ ú
ë 100û ë 100û
LF I O
MG R Jn P

CI = Amount – Principal =PMH1+ K -1P
N 100 Q

When Rates are different for different years, say R , R , R % for 1st, 2nd & 3rd years respectively, then,
1 2 3
é R ùé R ùé R ù

Amount =P 1+ 1 1+ 2 1+ 3
ê úê úê ú
ë 100ûë 100ûë 100û
In general, interest is considered to be SIMPLE unless otherwise stated.
GEOMETRY

(i) Sum of all the exterior angle of a polygon = 360°
360°